Show that the statement

p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by

(i) direct method

(iii) method of contrapositive

Asked by Pragya Singh | 1 year ago |  276

##### Solution :-

Let p: ‘If x is a real number such that x3 + 4x = 0, then x is 0’

q: x is a real number such that x3 + 4x = 0

r: x is 0

(i) We assume that q is true to show that statement p is true and then show that r is true

Therefore, let statement q be true

Hence, x3 + 4x = 0

x (x2 + 4) = 0

x = 0 or x2 + 4 = 0

Since x is real, it is 0.

So, statement r is true.

Hence, the given statement is true.

(ii) By contradiction, to show statement p to be true, we assume that p is not true.

Let x be a real number such that x3 + 4x = 0 and let x ≠ 0

Hence, x3 + 4x = 0

x (x2 + 4) = 0

x = 0 or x2 + 4 = 0

x = 0 or x2 = -4

However x is real. Hence, x = 0, which is a contradiction since we have assumed that x ≠ 0

Therefore, the given statement p is true.

(iii) By contrapositive method, to prove statement p to be true, we assume that r is false and prove that q must be false

∼r: x ≠ 0

Clearly, it can be seen that

(x2 + 4) will always be positive

x ≠ 0 implies that the product of any positive real number with x is not zero.

Now, consider the product of x with (x2 + 4)

∴ x (x2 + 4) ≠ 0

x3 + 4x ≠ 0

This shows that statement q is not true.

Hence, proved that

∼r ⇒ ∼q

Hence, the given statement p is true.

Answered by Abhisek | 1 year ago

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